The conductivity of 0.001028 mol L-1 acetic acid is 4.95 × 10-5 S cm-1. Calculate its dissociation constant if ∧m∘ for acetic acid is 390.5 S cm2 mol-1.

Question

The conductivity of 0.001028 mol L-1 acetic acid is 4.95 × 10-5 S cm-1. Calculate its dissociation constant if ∧m∘ for acetic acid is 390.5 S cm2 mol-1. (A) 1.78 × 10-5 mol L-1 (B) 1.87×10-5 mol L-1 (C) 0.178×10-5 mol L-1 (D) 0.0178×10-5 mol L-1

Tardigrade Admin · Accepted Answer

Lambdam =(k/c)=(4.95×10 -5S cm-1/0.001028 mol L-1) ×(1000 cm3/L) =48.15 S cm2 mol-1 α=( Lambdam/ Lambdamâ)=(48.15/390.5)=0.1233 k=(c α2/(1-α))=(0.001028×(0.1233)2/1-0.1233)=1.78×10-5 mol L-1

Q. The conductivity of 0.001028 mol $L^{- 1}$ acetic acid is 4.95 × $1 0^{- 5} S c m^{- 1} .$ Calculate its dissociation constant if $\land_{m}^{\circ}$ for acetic acid is $390.5 S c m^{2} m o l^{- 1} .$

$1.78 \times 1 0^{- 5} m o l L^{- 1}$

$1.87 \times 1 0^{- 5} m o l L^{- 1}$

$0.178 \times 1 0^{- 5} m o l L^{- 1}$

$0.0178 \times 1 0^{- 5} m o l L^{- 1}$

Q. The conductivity of 0.001028 mol L−1 acetic acid is 4.95 × 10−5Scm−1. Calculate its dissociation constant if ∧m∘​ for acetic acid is 390.5Scm2mol−1.

1.78×10−5molL−1

1.87×10−5molL−1

0.178×10−5molL−1

0.0178×10−5molL−1

Λm​=ck​=0.001028molL−14.95×10−5Scm−1​×L1000cm3​ =48.15Scm2mol−1 α=Λm∘​Λm​​=390.548.15​=0.1233 k=(1−α)cα2​=1−0.12330.001028×(0.1233)2​=1.78×10−5molL−1

Q. The conductivity of 0.001028 mol $L^{- 1}$ acetic acid is 4.95 × $1 0^{- 5} S c m^{- 1} .$ Calculate its dissociation constant if $\land_{m}^{\circ}$ for acetic acid is $390.5 S c m^{2} m o l^{- 1} .$

$1.78 \times 1 0^{- 5} m o l L^{- 1}$

$1.87 \times 1 0^{- 5} m o l L^{- 1}$

$0.178 \times 1 0^{- 5} m o l L^{- 1}$

$0.0178 \times 1 0^{- 5} m o l L^{- 1}$